Compound 'A' was prepared by oxidation of compound 'B' with alkaline $KMnO_4$. Compound 'A' on reduction with lithium aluminium hydride gets converted back to compound 'B'. When compound 'A' is heated with compound 'B' in the presence of $H_2SO_4$ it produces a fruity smell of compound 'C'. To which family do the compounds 'A', 'B' and 'C' belong?

> [!TLDR] > The key is the interconversion between an aldehyde/ketone and an alcohol via oxidation and reduction, and the esterification reaction that produces a fruity smell. Compound **A** is a **carboxylic acid**, **B** is an **alcohol**, and **C** is an **ester**. The problem describes a cycle of reactions that is a classic fingerprint of organic functional group transformations. Let’s decode each clue. 1. **Oxidation of B with alkaline KMnO₄ gives A.** Alkaline KMnO₄ is a strong oxidizing agent. It typically oxidizes primary alcohols to carboxylic acids and secondary alcohols to ketones. Tertiary alcohols are not oxidized under these conditions. So B must be an alcohol (primary or secondary), and A is either a carboxylic acid or a ketone. 2. **Reduction of A with LiAlH₄ gives back B.** Lithium aluminium hydride (LiAlH₄) is a powerful reducing agent that reduces carboxylic acids to primary alcohols and ketones to secondary alcohols. This tells us that the reduction of A regenerates B exactly. So the oxidation-reduction pair is reversible: B → A (oxidation) and A → B (reduction). This is only possible if B is an alcohol and A is the corresponding carbonyl compound (aldehyde/ketone or carboxylic acid). But note: if A were an aldehyde, it would be further oxidized to a carboxylic acid by KMnO₄, so A cannot be an aldehyde here. So A is either a carboxylic acid (from primary alcohol) or a ketone (from secondary alcohol). 3. **Heating A with B in presence of H₂SO₄ gives a fruity-smelling compound C.** A fruity smell is characteristic of esters. Esters are formed when a carboxylic acid reacts with an alcohol in the presence of an acid catalyst (Fischer esterification). So this reaction requires A to be a carboxylic acid and B to be an alcohol. If A were a ketone, it would not form an ester with an alcohol under these conditions. Therefore, A must be a carboxylic acid, and B must be an alcohol. …

Compound 'A' was prepared by oxidation of compound 'B' with alkaline KMnO₄. Compound 'A' on reduction with lit

NCERT Exemplar · Q23

KMnO_4

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Concept understanding — Oxidation Reactions

Alcohol Oxidation: The Intuition First

Imagine you have a molecule of ethanol — the alcohol in your hand sanitizer or a drink. It has a carbon atom bonded to an –OH group. Now picture that –OH group as a "handle" that can be transformed. Oxidation, in organic chemistry, doesn't always mean adding oxygen — it often means removing hydrogen from a carbon that already has a bond to oxygen. For alcohols, oxidation is like "stripping away" hydrogen atoms from the carbon that holds the –OH, turning the alcohol into a more oxidized functional group.

Think of it this way: a primary alcohol (R–CH₂–OH) has two hydrogens on the carbon with the –OH. If you remove one hydrogen and the hydrogen from the –OH, you get an aldehyde (R–CHO). Remove both hydrogens (and the –OH hydrogen), and you get a carboxylic acid (R–COOH). A secondary alcohol (R–CHOH–R') has only one hydrogen on that carbon — remove it, and you get a ketone (R–CO–R'). A tertiary alcohol has no hydrogen on that carbon — so it cannot be oxidized without breaking the carbon skeleton.

That's the core intuition: oxidation of an alcohol is about removing hydrogens from the carbon bearing the –OH group. The more hydrogens you can remove, the more oxidized the product.

The Precise Statement

Alcohol oxidation is the process in which an alcohol loses hydrogen atoms (dehydrogenation) from the carbon bonded to the -OH group, increasing the number of C-O bonds (or decreasing C-H bonds). The outcome depends on the class of the alcohol:

Alcohol Class	Structure	Product after oxidation	Reagent example
Primary (1°)	R–CH₂–OH	Aldehyde (R–CHO) then Carboxylic acid (R–COOH)	PCC (stops at aldehyde); K₂Cr₂O₇/H⁺ (goes to acid)
Secondary (2°)	R–CHOH–R'	Ketone (R–CO–R')	K₂Cr₂O₇/H⁺, CrO₃, etc.
Tertiary (3°)	R₃C–OH	No reaction (under normal conditions)	—

Watch out

A common mistake: students think "oxidation" always adds oxygen. For alcohols, it's removal of hydrogen from the carbon with the –OH. The oxygen from the –OH stays — it's the hydrogens that leave.

Why Does Tertiary Alcohol Not Oxidize?

Look at the carbon with the –OH in a tertiary alcohol: it has three carbon groups attached and no hydrogen. To form a C=O bond, you'd need to remove a hydrogen from that carbon — but there is none. The only way to oxidize a tertiary alcohol is to break a C–C bond (strong and difficult), which is not typical oxidation. So in standard organic chemistry, tertiary alcohols are inert to mild oxidizing agents.

A Real-World Analogy

Think of the alcohol carbon as a "parking spot" with a certain number of hydrogen "cars." Primary alcohol has two cars parked. Oxidation is like towing away one car (→ aldehyde) or both cars (→ carboxylic acid). Secondary alcohol has one car — tow it away, and you get a ketone. Tertiary alcohol has zero cars — nothing to tow, so no reaction.

Key Reagents to Remember (for exams)

PCC (pyridinium chlorochromate): oxidizes 1° alcohols to aldehydes only — stops there.
K₂Cr₂O₇ / H₂SO₄ (Jones reagent): oxidizes 1° alcohols all the way to carboxylic acids; 2° alcohols to ketones.
KMnO₄: similar to dichromate, but stronger — can over-oxidize. …

Why this formula?

Oxidation Reactions: Why the Key Principles Hold

Oxidation reactions are fundamental to chemistry, and understanding why they work the way they do is essential for mastering Indian board exams (Class 11, 12, JEE, NEET). Let's break down the core ideas from first principles.

1. The Core Definition: What Does "Oxidation" Really Mean?

Historically, oxidation meant "adding oxygen." But that's too narrow. The modern, exam-correct definition is:

Oxidation is the loss of electrons by a species.

This is the electronic concept (given by the ionic theory). The why behind this definition comes from the behavior of atoms during chemical reactions.

Why do atoms lose electrons?

Atoms seek stability. They achieve this by having a full outer electron shell (octet, duplet, or pseudo-inert gas configuration).

Metals (like Na, Mg, Fe) have few valence electrons (1, 2, or 3). It's energetically easier for them to lose these electrons than to gain 5, 6, or 7.
Non-metals (like O, Cl, F) have many valence electrons (5, 6, or 7). It's energetically easier for them to gain electrons.

So, when a metal reacts with a non-metal, the metal loses electrons (gets oxidized), and the non-metal gains electrons (gets reduced).

Example:

2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}

Na loses 1 electron: $\text{Na} \rightarrow \text{Na}^+ + e^-$ (Oxidation)
Cl gains 1 electron: $\text{Cl}_2 + 2e^- \rightarrow 2\text{Cl}^-$ (Reduction)

Key takeaway: Oxidation and reduction always happen together (Redox reactions). You cannot have one without the other.

2. The Key Formula(e): Oxidation Number Rules

The oxidation number (O.N.) is a bookkeeping tool. It's not a real charge (except in ionic compounds), but it helps track electron flow.

Why do we assign oxidation numbers?

Because in covalent compounds (like $\text{CH}_4$ or $\text{H}_2\text{O}$ ), electrons are shared, not transferred. We need a way to pretend they are transferred to see which atom "owns" the electrons more.

The Rules (and why they exist)

Rule	Statement	Why this rule?
1	O.N. of an element in its free state = 0	No electron transfer has occurred.
2	O.N. of a monatomic ion = its charge	The atom has actually lost/gained that many electrons.
3	O.N. of H = +1 (except in metal hydrides where it's -1)	H is less electronegative than O, F, Cl, but more electronegative than metals.
4	O.N. of O = -2 (except in peroxides where it's -1, superoxides -1/2, and with F where it's +2)	O is highly electronegative (3.44 on Pauling scale). It "pulls" electrons toward itself.
5	Sum of O.N. in a neutral compound = 0	The compound has no net charge.
6	Sum of O.N. in a polyatomic ion = charge of the ion	The ion's overall charge must be accounted for.

The Derivation of a Key Formula: Finding O.N. of an Unknown Element

Suppose you need to find the O.N. of S in $\text{H}_2\text{SO}_4$ .

Step 1: Write known O.N.s:

H: +1 (rule 3)
O: -2 (rule 4)
S: let it be $x$ (unknown)

Step 2: Apply rule 5 (neutral compound sum = 0):

2(+1) + x + 4(-2) = 0

Step 3: Solve:

2 + x - 8 = 0

x - 6 = 0

x = +6

Why this works: The oxidation number is a mathematical consequence of the electronegativity hierarchy. Oxygen is more electronegative than sulfur, so it "takes" the electrons. Hydrogen is less electronegative than sulfur, so it "gives" electrons to sulfur. The net result is that sulfur appears to have lost 6 electrons.

3. The Key Formula(e): Balancing Redox Equations

Two methods are exam-critical: Oxidation Number Method and Ion-Electron Method (Half-Reaction Method).

Why do we need these methods?

Because in a redox reaction, the total number of electrons lost (oxidation) must equal the total number of electrons gained (reduction). This is the Law of Conservation of Charge.

The Ion-Electron Method (for acidic medium) — Step-by-step why

Example: Balance $\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$ (acidic)

Step 1: Write half-reactions.

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$

Why? Fe loses 1 electron (O.N. goes from +2 to +3).
Reduction: $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$

Why? Mn gains electrons (O.N. goes from +7 to +2).

Step 2: Balance atoms other than H and O.

Mn is already balanced (1 on each side).

Step 3: Balance O by adding $\text{H}_2\text{O}$ .

Left: 4 O atoms. Right: 0 O atoms.
Add 4 $\text{H}_2\text{O}$ to the right:

\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 4: Balance H by adding $\text{H}^+$ (because acidic medium).

Right: 8 H atoms (from 4 $\text{H}_2\text{O}$ ). Left: 0 H atoms.
Add 8 $\text{H}^+$ to the left:

8\text{H}^+ + \text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 5: Balance charge by adding electrons.

Left: $8(+1) + (-1) = +7$ charge.
Right: $+2$ charge.
To make left = right, add 5 electrons to the left:

8\text{H}^+ + \text{MnO}_4^- + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

Step 6: Multiply half-reactions to equalize electrons.

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (×5)
Reduction: $8\text{H}^+ + \text{MnO}_4^- + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ (×1)

Step 7: Add them:

5\text{Fe}^{2+} + 8\text{H}^+ + \text{MnO}_4^- \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O}

Why this works: Every step is driven by conservation laws:

Mass balance: Same number of each atom on both sides.
Charge balance: Net charge on left = net charge on right.
Electron balance: Electrons lost = electrons gained.

4. The Key Formula(e): Electrochemical Series and Cell Potential

For a galvanic cell (voltaic cell), the cell potential $E^\circ_{\text{cell}}$ is:

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

Why this formula? …

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The key is the interconversion between an aldehyde/ketone and an alcohol via oxidation and reduction, and the esterification reaction that produces a fruity smell. Compound A is a carboxylic acid, B is an alcohol, and C is an ester.

The problem describes a cycle of reactions that is a classic fingerprint of organic functional group transformations. Let’s decode each clue.

Oxidation of B with alkaline KMnO₄ gives A.

Alkaline KMnO₄ is a strong oxidizing agent. It typically oxidizes primary alcohols to carboxylic acids and secondary alcohols to ketones. Tertiary alcohols are not oxidized under these conditions. So B must be an alcohol (primary or secondary), and A is either a carboxylic acid or a ketone.
Reduction of A with LiAlH₄ gives back B.

Lithium aluminium hydride (LiAlH₄) is a powerful reducing agent that reduces carboxylic acids to primary alcohols and ketones to secondary alcohols. This tells us that the reduction of A regenerates B exactly. So the oxidation-reduction pair is reversible: B → A (oxidation) and A → B (reduction). This is only possible if B is an alcohol and A is the corresponding carbonyl compound (aldehyde/ketone or carboxylic acid). But note: if A were an aldehyde, it would be further oxidized to a carboxylic acid by KMnO₄, so A cannot be an aldehyde here. So A is either a carboxylic acid (from primary alcohol) or a ketone (from secondary alcohol).
Heating A with B in presence of H₂SO₄ gives a fruity-smelling compound C.

A fruity smell is characteristic of esters. Esters are formed when a carboxylic acid reacts with an alcohol in the presence of an acid catalyst (Fischer esterification). So this reaction requires A to be a carboxylic acid and B to be an alcohol. If A were a ketone, it would not form an ester with an alcohol under these conditions. Therefore, A must be a carboxylic acid, and B must be an alcohol. …

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Here are the common mistakes students make on this classic interconversion problem, along with the correct reasoning to avoid them.

Mistake 1: Misidentifying the Oxidation Step

The Mistake: Students often think alkaline $\text{KMnO}_4$ is a strong oxidant that will cleave carbon-carbon bonds or produce $\text{CO}_2$ . They then guess 'A' is a ketone or an aldehyde.
Why it's wrong: Alkaline $\text{KMnO}_4$ (cold, dilute) is a mild oxidant. It specifically oxidizes primary alcohols to carboxylic acids and secondary alcohols to ketones. It does not break C-C bonds here.
How to avoid: Memorize the specificity:
- Primary alcohol $\xrightarrow{\text{alk. KMnO}_4}$ Carboxylic acid
- Secondary alcohol $\xrightarrow{\text{alk. KMnO}_4}$ Ketone
- Tertiary alcohol $\rightarrow$ No reaction.

Mistake 2: Forgetting the Reverse Reaction (Reduction)

The Mistake: Students correctly identify 'A' as a carboxylic acid but then forget that $\text{LiAlH}_4$ reduces a carboxylic acid all the way back to a primary alcohol, not to an aldehyde.
Why it's wrong: $\text{LiAlH}_4$ $LiAlH_{4}$ is a strong reducing agent. It reduces:
- Carboxylic acid $\rightarrow$ Primary alcohol
- Ketone $\rightarrow$ Secondary alcohol
- Ester $\rightarrow$ Two alcohols If 'A' were an aldehyde, $\text{LiAlH}_4$ would still give a primary alcohol, but the oxidation step wouldn't match.
How to avoid: Write the reduction products for each possibility. The only pair that works is:
- B = Alcohol
- A = Carboxylic acid (or ketone)
- Reduction of A gives back B.

Mistake 3: Ignoring the "Fruity Smell" Clue

The Mistake: Students guess 'C' is an aldehyde or a ketone because they associate "fruity smell" with any organic compound.
Why it's wrong: The reaction described is esterification:

\text{Carboxylic acid (A)} + \text{Alcohol (B)} \xrightarrow{\text{H}_2\text{SO}_4} \text{Ester (C)} + \text{H}_2\text{O}

Esters are famous for their fruity smells (e.g., ethyl acetate smells like pear).

How to avoid: Link "fruity smell" directly to esters. If you see an acid + alcohol + $\text{H}_2\text{SO}_4$ (heat), the product is always an ester.

Mistake 4: Confusing the Family Names

The Mistake: Students write "A = carboxylic acid, B = alcohol, C = ester" but then say "A belongs to the alcohol family" or "C belongs to the acid family."
Why it's wrong: Each compound belongs to a distinct functional group family:
- A = Carboxylic acid family (functional group: $-\text{COOH}$ ) …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-M1 markMCQ
$Q. Etard reaction is a method of preparation of benzaldehyde by oxidation of toluene. The oxidizing agent used in this reaction is: (A) Chromic oxide (B) Chromyl chloride (C) Potassium dichromate (D) Pyridinium chlorochromate$
›Reveal solutionSolution
The Etard reaction uses chromyl chloride ( $\text{CrO}_2\text{Cl}_2$ ) to selectively oxidize the methyl group of toluene to an aldehyde, stopping at benzaldehyde without overoxidation. The correct option is (B).

The Etard reaction is a classic, elegant method for converting a methyl group attached to an aromatic ring (like toluene) directly into an aldehyde group (like benzaldehyde). The key challenge in such oxidations is stopping at the aldehyde stage — most strong oxidizers (like potassium dichromate in acid) would push all the way to benzoic acid. The genius of the Etard reaction lies in using a specific, milder oxidizing agent that forms a stable intermediate complex, preventing overoxidation.

Why chromyl chloride?

Chromyl chloride ( $\text{CrO}_2\text{Cl}_2$ ) is a powerful but selective oxidant. It reacts with the benzylic C–H bonds of toluene to form a solid, insoluble complex (often called the Etard complex). This complex can be isolated and then hydrolyzed (with water or dilute acid) to release benzaldehyde. The chromium is reduced, but the aldehyde is protected within the complex until workup. Other chromium(VI) reagents like chromic oxide or potassium dichromate are too aggressive in acidic media — they generate the aldehyde but immediately oxidize it further. Pyridinium chlorochromate (PCC) is milder but typically used in anhydrous conditions for alcohols, not for direct methyl-to-aldehyde conversion on toluene.

Let’s walk through the reasoning step by step.

Identify the goal: We need an oxidant that converts the methyl group ( $\text{–CH}_3$ ) of toluene to a formyl group ( $\text{–CHO}$ ) without further oxidation to a carboxyl group ( $\text{–COOH}$ ). This requires a reagent that either (a) forms a protective intermediate or (b) is inherently mild enough to stop at the aldehyde.

Evaluate option (A) – Chromic oxide ( $\text{CrO}_3$ ): In aqueous acidic conditions, chromic oxide is a very strong oxidizer. It would oxidize toluene first to benzyl alcohol, then to benzaldehyde, and then rapidly to benzoic acid. It does not form a stable isolable intermediate with the aldehyde. So this is not suitable for stopping at benzaldehyde.

Evaluate option (B) – Chromyl chloride ( $\text{CrO}_2\text{Cl}_2$ ): This is the classic Etard reagent. It reacts with toluene in carbon disulfide or carbon tetrachloride to form a brownish-red precipitate — the Etard complex. The complex is thought to be a cyclic adduct where chromium is coordinated to the benzylic carbon and oxygen. Upon hydrolysis, this complex decomposes to give benzaldehyde. The key is that the aldehyde is “masked” in the complex until workup, preventing overoxidation. This is the correct reagent. …
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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{KMnO}_4$
›Reveal solutionSolution
The key idea is that hot acidic KMnO₄ cleaves alkenes at the double bond, turning each doubly bonded carbon into a carbonyl group (ketone or carboxylic acid). The products given — ethanoic acid and propanone — uniquely point to the alkene 2‑methylbut‑2‑ene, option (B).

Concept & Intuition

Hot acidic potassium permanganate (KMnO₄) is a strong oxidising agent. When it reacts with an alkene, it doesn’t just stop at a diol — it cleaves the carbon‑carbon double bond completely. Each carbon that was part of the double bond becomes a separate carbonyl compound:

If that carbon had two alkyl groups attached, it becomes a ketone.

If it had one alkyl group and one hydrogen, it becomes a carboxylic acid (which may further oxidise to CO₂ if it’s a terminal carbon, but here we have stable products).

If it had two hydrogens (terminal =CH₂), it becomes CO₂ and water.

So the puzzle is: Which alkene, when cut at the double bond, gives exactly one molecule of ethanoic acid (CH₃COOH) and one molecule of propanone (CH₃COCH₃)?

Step‑by‑step reasoning

Identify the fragments from the products Ethanoic acid is CH₃–COOH. That means one half of the original double bond must have been a carbon with a methyl group and a hydrogen:

$\text{CH}_3\text{–CH}= \quad \text{(the =C becomes –COOH after oxidation)}$
Propanone is CH₃–CO–CH₃. That means the other half of the double bond must have been a carbon with two methyl groups attached:
$\text{–C(CH}_3)_2 \quad \text{(the =C becomes >C=O)}$

Reconstruct the alkene Join the two fragments at the double bond:

$\text{CH}_3\text{–CH} = \text{C(CH}_3)_2$
This is 2‑methylbut‑2‑ene.

(Check the name: a 4‑carbon chain with a double bond between C2 and C3, and a methyl substituent on C2.)

Verify with the options

(A) Pent‑2‑ene: CH₃–CH=CH–CH₂–CH₃ → would give propanoic acid + ethanoic acid (not propanone). …
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COMEDK 2025Set 2025-E1 markMCQ
$[\mathrm{X}]$
›Reveal solutionSolution
$[\text{X}]$ hydrolyses to an alcohol $[\text{B}]$ and acid $[\text{C}]$ ; since $[\text{B}]$ is oxidised (Jones reagent) to the same acid $[\text{C}]$ , and $[\text{C}]$ forms an anhydride with $\text{P}_2\text{O}_5$ , $[\text{X}]$ is propyl propanoate, option (A).

Work through the clues.

$[\text{X}]$ ( $\text{C}_6\text{H}_{12}\text{O}_2$ ) $\xrightarrow{\text{dil. }\text{H}_2\text{SO}_4}$ alcohol $[\text{B}]$ $+$ carboxylic acid $[\text{C}]$ — so $[\text{X}]$ is an ester.

$[\text{B}] \xrightarrow{\text{Jones reagent}} [\text{C}]$ : the alcohol is oxidised to the acid. A primary alcohol $\text{R–CH}_2\text{OH}$ oxidises to $\text{R–COOH}$ . For this product to be $[\text{C}]$ itself, the acid and alcohol must share the same carbon skeleton — i.e. the ester is $\text{R–COO–CH}_2\text{–R}$ with both halves derived from the same acid.

$[\text{C}] \xrightarrow{\text{P}_2\text{O}_5,\ \Delta}$ anhydride, confirming $[\text{C}]$ is a carboxylic acid.

Test option (A), $\text{C}_2\text{H}_5\text{–COO–(CH}_2)_2\text{–CH}_3$ (propyl propanoate):

Hydrolysis $\rightarrow$ propan-1-ol $[\text{B}]$ $+$ propanoic acid $[\text{C}]$ . …
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KCET 2022Set B-31 markMCQ
$Q. The test to differentiate between pentan-2-one and pentan-3-one is (A) Fehling’s test (B) Iodoform test (C) Baeyer’s test (D) Benedict’s test$
›Reveal solutionSolution
Only a methyl ketone gives iodoform; pentan-2-one is one and pentan-3-one is not, so $I_2$ /NaOH separates the pair.

1. The two compounds
$\text{Pentan-2-one: } CH_3\!-\!\overset{O}{\overset{||}{C}}\!-\!CH_2\!-\!CH_2\!-\!CH_3$ $\text{Pentan-3-one: } CH_3\!-\!CH_2\!-\!\overset{O}{\overset{||}{C}}\!-\!CH_2\!-\!CH_3$
They are positional isomers ( $C_5H_{10}O$ ) — same functional group, different position. So any test that merely detects a ketone will respond identically to both. We need a test sensitive to the position of the carbonyl.

2. Why the iodoform test discriminates

The iodoform reaction requires a $CH_3$ group directly attached to the carbonyl carbon (a methyl ketone), or a $CH_3CH(OH)-$ group that can be oxidised to one. Mechanism: the three $\alpha$ -H's of that methyl group are successively replaced by iodine under basic conditions; the resulting $-CI_3$ is an excellent leaving group and is expelled by hydroxide as $CHI_3$ — a yellow crystalline precipitate with a characteristic antiseptic smell.
$CH_3COR + 3I_2 + 4NaOH \longrightarrow CHI_3\!\downarrow + RCOONa + 3NaI + 3H_2O$

Pentan-2-one: the carbonyl carries a $CH_3$ group $\Rightarrow$ positive (yellow $CHI_3$ ; the other product is sodium butanoate). …
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KCET 2019Set A-11 markMCQ
$NaHCO_3$
›Reveal solutionSolution

Ethanol is too weak an acid for litmus or carbonate tests; only sodium metal is basic/reactive enough to pull off its $-OH$ proton, releasing $H_2$ .

Step 1 — How weak is ethanol as an acid?

Ethanol ionises as $C_2H_5OH \rightleftharpoons C_2H_5O^- + H^+$ , with $pK_a\approx 16$ — it is even weaker than water ( $pK_a\approx14$ ), and far weaker than carbonic acid ( $pK_a\approx6.4$ ) or a carboxylic acid ( $pK_a\approx5$ ). The alkyl group is electron-donating (+I effect), which destabilises the alkoxide ion $C_2H_5O^-$ and suppresses ionisation.

Step 2 — Why (A) fails.

Blue litmus turns red only for an acid strong enough to give an appreciable $[H^+]$ in water. Ethanol is neutral to litmus — a classic exam point.

Step 3 — Why (B) and (C) fail.

$NaHCO_3$ and $Na_2CO_3$ liberate $CO_2$ only with acids stronger than carbonic acid — that is the standard test that distinguishes carboxylic acids (which do effervesce) from alcohols and phenols (which do not). Ethanol, at $pK_a\approx16$ , is nowhere near strong enough, so no brisk effervescence occurs. (Even phenol, $pK_a\approx10$ , fails this test.) …
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KCET 2018Set A-11 markMCQ
$\ce{CH3} \quad \ce{CrO2Cl2} \quad \ce{H3O^+}$
›Reveal solutionSolution

$\ce{CrO2Cl2}$ in $\ce{CS2}$ (Etard reaction) oxidises toluene's methyl group only to the aldehyde stage — the intermediate chromium complex X hydrolyses to benzaldehyde, Z.

Step 1 — Recognise the reagent.

$\ce{CrO2Cl2}$ is chromyl chloride; used in an inert solvent such as $\ce{CS2}$ or $\ce{CCl4}$ on a methyl arene, this is the classic Etard reaction.

Step 2 — What X is.

Chromyl chloride attacks the benzylic $\ce{-CH3}$ group and forms a brown chromium complex (an addition complex at the benzylic carbon):
$\ce{C6H5-CH3 + CrO2Cl2 ->[CS2] C6H5-CH(OCrOHCl2)2}\quad (\text{= X, the Etard complex})$
The key point is that the oxidation is arrested at this complex — the carbon is not oxidised further while it is tied up in the complex.

Step 3 — What Z is.

Hydrolysis of the Etard complex with $\ce{H3O+}$ liberates the carbonyl compound at the aldehyde oxidation level:

$\ce{C6H5-CH(OCrOHCl2)2 ->[H3O+] C6H5-CHO}$ …
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Concept understanding — Oxidation Reactions

Alcohol Oxidation: The Intuition First

The Precise Statement

Why Does Tertiary Alcohol Not Oxidize?

A Real-World Analogy

Key Reagents to Remember (for exams)

Why this formula?

Oxidation Reactions: Why the Key Principles Hold

1. The Core Definition: What Does "Oxidation" Really Mean?

Why do atoms lose electrons?

2. The Key Formula(e): Oxidation Number Rules

Why do we assign oxidation numbers?

The Rules (and why they exist)

The Derivation of a Key Formula: Finding O.N. of an Unknown Element

3. The Key Formula(e): Balancing Redox Equations

Why do we need these methods?

The Ion-Electron Method (for acidic medium) — Step-by-step why

4. The Key Formula(e): Electrochemical Series and Cell Potential

Why this formula? …

Concept: Interconversion of Carboxylic Acids, Alcohols, and Esters

Method: Functional Group Mapping via Reagent Action

Mistake 1: Misidentifying the Oxidation Step

Mistake 2: Forgetting the Reverse Reaction (Reduction)

Mistake 3: Ignoring the "Fruity Smell" Clue

Mistake 4: Confusing the Family Names

1. The two compounds

2. Why the iodoform test discriminates

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